Question

Como resolver? PS: quero saber o COS de Alfa $<var>(2\sqrt{3})^{2} = (3\sqrt{2})^{2} + (\sqrt{3} + 3)^{2} - 2.3\sqrt{2}.\sqrt{3}+3.Cos\alpha</var>$

Answer 1

$2\sqrt{3} ^{2} = (3\sqrt{2})^{2}+ (\sqrt{3}+3) ^{2} - 2*3 \sqrt{2} * \sqrt{3} +3*cos\alpha$
$((2+2)\sqrt{3}) = (3\sqrt{2})^{2}+ (\sqrt{3}+3) ^{2} - 2*3 \sqrt{2} * \sqrt{3} +3*cos \alpha$
$4\sqrt{3}= (3\sqrt{2})^{2}+ (\sqrt{3}+3) ^{2} - 2*3 \sqrt{2} * \sqrt{3} +3*cos \alpha$

$4\sqrt{3} = ((3+3)\sqrt{2})+ (\sqrt{3}+3) ^{2} - 2*3 \sqrt{2} * \sqrt{3} +3*cos \alpha$ 

$4\sqrt{3} = ((6)\sqrt{2})+ (\sqrt{3}+3) ^{2} - 2*3 \sqrt{2} * \sqrt{3} +3*cos \alpha$

$4\sqrt{3} = ((6)\sqrt{2})+ (11+6\sqrt{3}) - 2*3 \sqrt{2} * \sqrt{3} +3*cos \alpha$

$4\sqrt{3} = (6\sqrt{2})+ (11+6\sqrt{3}) - 6\sqrt{6} +3*cos \alpha$

$\frac{ - 2\sqrt{3} - 6(\sqrt{2} +\sqrt{6}) - 11 }{3} = cos \alpha$